Problem: What is the slope of the line tangent to $f(x) = 2x^{2}+3x-7$ at $x = 1$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(2(x+h)^{2}+3(x+h)-7) - (2x^{2}+3x-7)}{h}$ $ = \lim_{h \to 0} \frac{(2(x^{2}+2x h+h^{2})+3(x+h)-7) - (2x^{2}+3x-7)}{h}$ $ = \lim_{h \to 0} \frac{2x^{2}+4(x h)+2h^{2}+3x+3h-7-2x^{2}-3x+7}{h}$ $ = \lim_{h \to 0} \frac{4(x h)+2h^{2}+3h}{h}$ $ = \lim_{h \to 0} 4x+2h+3$ $ = 4x+3$ $ = (4)(1)+3$ $ = 7$